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allochka39001 [22]
3 years ago
5

During a storm, the power went out at Bailey’s house. She had no electricity. What could she use as an alternate form of electri

cal energy?
A) candles
B) a fire in the fireplace
C) a battery-operated lamp
D) a light in the refrigerator
Physics
2 answers:
Alisiya [41]3 years ago
6 0
A . Can be renewable and last a long time
evablogger [386]3 years ago
4 0
C a battery-operated lamp sorry if incorrect
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Arocket with an initial velocity of 20 m/s another engine that gives it an acceleration of 4 m/s ^ 2 over 10 secondsHow far did
ExtremeBDS [4]

Answer:

50m

   

Explanation:

Given parameters:

Initial velocity  = 20m/s

Acceleration  = 4m/s²

Time  = 10s

Unknown:

Distance traveled by the rocket  = ?

Solution:

To solve this problem use the expression below;

   

     v² = u² + 2as

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

  final velocity = 0

 Insert the parameters and solve;

  0²  = 20² + 2 x 4 x s

   -400  = 8s

         s  = 50m

   Disregard the negative sign because distance cannot be negative.

3 0
2 years ago
a basketball player can leap upward .65m how long does the basketball player remain in the air use 9.81m/s²​
tigry1 [53]

At the player's maximum height, their velocity is 0. Recall that

{v_f}^2-{v_i}^2=2a\Delta y

which tells us the player's initial velocity v_i is

0^2-{v_i}^2=-2g(0.65\,\mathrm m)\implies v_i=3.6\dfrac{\rm m}{\rm s}

The player's height at time t is given by

y=v_it-\dfrac g2t^2

so we find their airtime to be

0.65\,\mathrm m=\left(3.6\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2\implies t=0.36\,\mathrm s

6 0
3 years ago
The Earth orbits the Sun at a speed of 30 km/s. At that speed it completes one path around the Sun every year. Of course, as tha
romanna [79]

Answer:

a. 299,792,458 m/s

Explanation:

Since the speed of light in a vacuum is invariant and has the value of 299,792,458 m/s, we would measure this value of 299,792,458 m/s for the speed of light from the star as it arrives on Earth.

3 0
3 years ago
Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w
Tatiana [17]

Answer:

The average velocity is

266\frac{m}{s},274\frac{m}{s} and 117\frac{m}{s} respectively.

Explanation:

Let's start writing the vertical position equation :

L(t)=2t^{3}+t^{2}-5t+1

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

V_{avg}=\frac{Displacement}{Time} = Δx / Δt = \frac{x2-x1}{t2-t1}

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

t2-t1=8s-5s=3s

For the position variation we use the vertical position equation :

x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m

x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

\frac{798m}{3s}=266\frac{m}{s}

For the second time interval :

t1 = 4 s → t2 = 9 s

x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m

x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

\frac{1370m}{5s}=274\frac{m}{s}

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m

x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

\frac{702m}{6s}=117\frac{m}{s}

5 0
3 years ago
Different _ of an element have different numbers of neutrons?
natali 33 [55]

Answer:

Isotopes

Explanation:

Your welcome

4 0
3 years ago
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